7 Lecture 6, January 19, 2024
7.0.1 Multinomial Coefficient
Definition 7.1 (Multinomial Coefficient) Consider \(n\) objects which consist of \(k\) types. Suppose that there are \(n_1\) objects which are of type 1, \(n_2\) which are of type 2, and in general \(n_i\) objects of type \(i\). Then there are \[ \frac{n!}{n_1 ! n_2! \dots n_k !} \] distinguishable arrangements of the \(n\) objects. This quantity is known as a multinomial coefficient and denoted by \[ \binom{n}{n_1,n_2,\dots,n_k}= \frac{n !}{n_1 ! n_2! \dots n_k !}. \]
Note: Multinomial coefficient is an extension of the binomial coefficient. In binomial coefficient, there are only two groups/objects, and the first type has size \(n_1\) and the size of the second type is consequently \(n-n_1\), where \(n\) is the total number of objects. Hence we have \({n \choose n_1} = \frac{n!}{n_1! (n-n_1)!}\). Try to compare this with the multinomial coefficient.
7.0.2 The Birthday Problem
Suppose a room contains \(n\) people. What is the probability at least two people in the room share a birthday?
Assumption: Suppose that each of the \(n\) people is equally likely to have any of the 365 days of the year as their birthday, so that all possible combinations of birthdays are equally likely.
Let \(A\) be the event that at least two people share a birthday. Then \[ P(A) = 1 - P(A^c),\] where \(A^c\) is the event that nobody shares birthday with each other.
For \(n\) people to have unique birthdays, we need to arrange them among 365 days w/o replacement. Thus, \[|A^c| = 365^{(n)}.\]
For the size of the sample space, we see that each person has 365 possibilities for their birthday. Thus, \[|S| = 365^n.\]
Since we are assuming that all possible combinations of birthdays are equally likely, our desired probability becomes \[ P(A) = 1 - P(A^c) = 1 - \frac{365^{(n)}}{365^n} = 1 - \frac{n! {365 \choose n}}{365^n}. \]
For \(n\in\{100, 30, 23\}\) we find \[P(A_{100})= .9999997,\;\;\; P(A_{30})=.7063 \;\;\;\; P(A_{23})=.5073.\]